# Is there a compact subset of the irrationals with positive Lebesgue measure?

Does there exist $K \subseteq \mathbb{R} \backslash \mathbb{Q}$ such that $K$ is compact, and has Lebesgue measure greater than $0$? As I have been trying to think of examples, I suspect that any subset of $\mathbb{R} \backslash \mathbb{Q}$ that is closed can be at most countable, since the closure of an uncountable subset of irrationals should contain some rationals. And, the Lebesgue measure of a countable set is $0$. If there are any examples of such a set, I would be very interested to know how it is constructed.

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A modified fat Cantor set (en.wikipedia.org/wiki/Smith%E2%80%93Volterra%E2%80%93Cantor_set) might do it. –  Qiaochu Yuan Feb 9 '11 at 1:16
Inner regularity immediately implies this. –  Andres Caicedo Mar 27 '14 at 23:59

The answer is yes. Count the rationals in $[0,1]$ as $r_1,r_2,\ldots$, let $I_k$ be an open interval containing $r_k$ of length $3^{-k}$, and let $K=[0,1]\setminus\cup_k I_k$.
Come to think of it, "most straightforward" depends on how Lebesgue measure is constructed. The way I first learned it is in terms of defining outer measure using covering sequences of intervals and then imposing the Carathéodory criterion (or equivalent) for measurability. However, in Rudin's Real and complex analysis, Lebesgue measure comes from an application of the Riesz representation theorem, and its regularity follows by an application of a general theorem on the locally compact & $\sigma$-compact case. –  Jonas Meyer Feb 9 '11 at 1:49